How do you differentiate #y=xcosy^2-xy#?

Answer 1

#dy/dx=(cosy^2-y)/(1+2xysiny^2+x)#

#y=xcosy^2-xy#

We can differentiate each component and then put it together

#y'=dy/dx#

The next one is a chain differentiation and differentiation of a product

#(xcosy^2)'=1*cosy^2 +x(-siny^2)2ydy/dx# #= cosy^2-(2xysiny^2)dy/dx#
The next one is #(xy)'=1*y+xdy/dx#

Putting it together

#dy/dx=cosy^2-(2xysiny^2)dy/dx-y-xdy/dx#
#dy/dx(1+2xysiny^2+x)=cosy^2-y#
#dy/dx=(cosy^2-y)/(1+2xysiny^2+x)#
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Answer 2

To differentiate ( y = x \cos(y^2) - xy ), you would use the product rule and chain rule. The derivative with respect to ( x ) is:

[ \frac{dy}{dx} = \cos(y^2) - 2xy \sin(y^2) - y ]

This comes from differentiating ( x \cos(y^2) ) and ( -xy ) separately.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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